M03-09

Distance between Washington (W) and Baltimore (B) = 120
Relative speed of trains A and B = 30+90=120
Therefore, the trains together cover a distance of 120 miles in 1 hr and thus meet every hour.

At 1 hr, Train A travels 30 miles from W, Train B travels 90 miles from B. This is where the two trains meet (meeting pt 1).
At 2 hr, Train A reaches 60 miles from W, Train B travels 30 miles to W, turns around in negligible time and reaches 60 miles from W. This is where the two trains meet (meeting pt 2).

Distance between meeting pt 1 and pt2 = 60-30 = 30.

Answer B.

press kudos if the graphical representation helps you understand better!

Distance = 120 miles
A's speed : B's speed = 30:90 = 1:3
first meeting point -
A's distance = 1/4 x 120 = 30
B's distance = 3/4 x 120 = 90
So first meeting point is at 30 miles from Washington

At each meeting point, distance by A: distance by B will always be multiple of 1:3
When A=40 B = 120 (not meeting)
When A=50 B = 150 = -30 (not meeting : '-' implies change in direction)
When A =60 B = 180 = -60 (meeting)
So second meeting point is 60 miles from Washington.

Difference between meeting points is 60 -30 =30 miles

"Again Meeting" somewhat ambiguous. If it were written distance between the point where trains first meet and the point where faster train overtakes, it would be much easier. I calculated the point when trains "again meet" from opposite. And the answer is 60, which is also given.

The relative speed of the trains depends on direction - so when both trains are moving in opp directions it is 120mph, while the rel. speed when they move in the same direction is 60mph. so how did you reach the conclusion that they will keep meeting after every 1 hour.

How do we know they will be meeting every 1 hour?

Good question, shanks2020. I have taken the liberty of highlighting the original text that states as much. If we picture each train on a parallel track, shuttling back and forth from city to city (although who would ever want to take the slow train?), we can model the position of each train per hour to test the statement. I am going to use color to make the graphical interpretation a bit easier on the eyes. Each highlighted portion will represent a meeting of the two trains. I will also use an arrow (--> or <--) to indicate the number of 30-mile movements of each train per hour, as well as the direction of travel.

Train A (moves at 30 mph)
0---30--60---90---120 miles
|----|----|----|----|

Train B (moves at 90 mph)
120-90--60---30---0 miles
|----|----|----|----|

Again, the question states that the station turnaround times are negligible, so we can, for the sake of our simulation, assume constant motion. We must also ignore the rates at which each train would accelerate/decelerate, but with all of that said, let us take a look from the top:

Hour 1:
A: |----|----|----|----| -->
B: |----|----|----|----| <-- <-- <--

Hour 2:
A: |----|----|----|----| -->
B: |----|----|----|----| <-- --> -->

Hour 3:
A: |----|----|----|----| -->
B: |----|----|----|----| --> --> <--

Hour 4:
A: |----|----|----|----| -->
B: |----|----|----|----| <-- <-- <--

As you can see, after 4 hours of travel, the two trains will be in opposite cities at the same time, and it would not be until hour 5 that they would cross paths again and repeat the mirror of the hour 1 position:

Hour 5:
A: |----|----|----|----| <--
B: |----|----|----|----| --> --> -->

Thus, we can conclude that the trains will not meet every hour under the given conditions. Of course, this meta-analysis does not help at all with the problem at hand, but it does correct an assumption, the type that could get a test-taker into trouble on another (similar) question, and I hope it satisfies your curiosity.

Good luck with your studies, and thank you for opening the door to my light-hearted response. (I miss physics schematics.)

- Andrew

Even we have to do a similar analysis till the faster turns around.
Maybe Bunuel has a faster and direct way to tell something about the meeting points after the 1st meeting point. The least what i can think of is that we will have to find manually till the second meeting and then we can draw some pattern, based on the combined distance covered to meet once.

Hi shanks2020, I do not know what short-cut are you looking for. Bunuel has already given the most elegant solution. And MentorTutoring has explained how you visualise the solution.

There is a longer way to solve the problem, but you might take a few seconds more than 2 min to solve, hence not recommended. But this will help you understand the logic behind their solution.

So I am guessing you are okay with relative speeds so not going into that.
Step 1
Relative speed is 120mph and therefore they travel for 1 hour. Hence, the First meeting point is 30 miles away from Washington

Step 2
Now the faster train B has to reach the opposite station (Washington) at its respective speed (not relative speed here), as both trains have passed each other. B has 30 miles to traverse, so it takes =30/90 = 1/3 hour to do so. Now in this 1/3 hour the slower train A goes = 1/3 * 30 = 10 m further. Therefore, now the train A is 40 m away from Washington.

Step 3
Now, relative speed is 90-30 = 60mph. To traverse 40m in this speed trains meet again in = 40/60 = 2/3 hour
In 2/3 hour, A goes = 2/3 * 30 = 20 m

In step 2, A went 10 miles, and in step 3, A went 20 miles (shown by bold text). Therefore total distance away from the first meeting point = 30m

Hope this is clear.

Prices how?

I think Prices was talking about the real-life distance between the two cities, irrespective of the problem at hand. I used to live in Baltimore and traveled to D.C. from time to time. The driving distance via the Baltimore-Washington Parkway and MD-295 is under 40 miles, depending on where one lives in either place and where one is going. Regarding the GMAT™, I have seen some problems that use made-up names, perhaps to discourage any real-life knowledge from entering the picture (not that such knowledge would be useful), but other problems use actual locations, such as Chaco Canyon in the American southwest.

I am not sure any of this dialogue will lead to insights into the question at hand, but anyway, I guess my own life experience allowed me to understand exactly what Prices had meant.

- Andrew

I have a genuine doubt here. I have read a concept from a valid source that, if 2 bodies are travelling towards each other, then the total distance covered by both together for the nth meeting is (2n-1)D. Using that logic, here for the 3rd meeting total distance should for 2nd meeting should be 3*120 = 360. And ratio of distance covered would be 1:3 or 90 : 270. So slower body covers 90m when 2nd meet takes place. Hence, distance should have been 90-30 = 60.
But I get why this is not giving the correct answer: for a significant part both bodies are moving in same direction.
What i dont understand is, is there some general exception to the rule i have read then?

[quote="Bunuel"]

My method used drawing a circle resembling a clock with a distance of 120 marking every 1/4th of the circumference with a point at 12, 3, 6 and 9 o’clock. Place A and B depart from one of the points (let’s say at 12 o’clock position) in opposite directions and they meet at the 3 o’clock position in 1 hour. When A reaches point 12 o’clock it reverses direction and A and B meet each other at and then at another point at the 6 o’clock position. The distance from 3 and 6 o’clock position is 1/4 the circumference or 30.

It’s really the same method as the ones posted but just another way of visualizing the problem.

Using combined speed we get the speed =90mph+30mph=120mph
Total time to complete 120 miles with combined speed is 1hr. Distance traveled by first train in 1 hr is 30 miles and the second train in 1hr is 90 miles.

So, they meet first time when first train traveled 30mils
Similarly in 2nd hour they meet again at second point as using combined speed they have traveled 120 miles. Distance traveled by first train in 2nd hour is 60 miles.

Difference in distance between meeting points = 60-30 =30 miles

Posted from my mobile device

Let's say that everyone has easily figured out the first time both trains met (after 1h)

Now, assign x as the distance between the first and the second time they met, then x is the distance that the first train has gone between two meeting times.
Since the velocity of the second train is 3 times that of the first train, in the same amount of time, the distance that the second travels is 3 times the distance that the first travels.

The equation: 3x = 30x2 + x --> x=30

For those who want to explore more about the problem, try to change the velocity ( more "balance" between the two), and see what happens if in the second time they met, the first car has already reached Baltimore and then turned around. You will see that the aforementioned approach is still useful, with a little bit change in the equation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

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